limit of dirac delta and test functions I am trying to understand dirac delta better. We know that $\int_{\mathbb{R}^n} \delta(x) \phi(x)=\phi(0)$ where $\phi(x)$ is a test function and $\delta_{0}(x)$ represents dirac distribution. I have a function $\lim \limits_{l\rightarrow 0 }f(l)=c(constant)$. So, I think $\lim \limits_{l\rightarrow 0 }\int_{\mathbb{R}^n} \delta(x) f(l) \phi(x)=c\phi(0)$. What does happen, if I have such a function $\lim \limits_{l\rightarrow 0 }f_{l}(x)=\infty$, i.e like $f(x)=\frac{x}{l^2}$ 
 A: I think your notation is at the very least misleading: $\delta (x)$ makes little
sense, since it's a distribution, i.e. a functional on $\mathcal{D} (
\mathbb{R}^n)$. Although it's indeed quite common notation, handy for some
manipulations, I'd rather write $\delta ( \phi) = 0$ or $\langle \delta_0,
\phi \rangle_{\mathcal{D}' ( \mathbb{R}^n)} = \phi ( 0)$ if you like
verbosity. Now, if your function $f_l$ is in $\mathcal{D} (
\mathbb{R}^n) = C^{\infty}_0 ( \mathbb{R}^n)$, you can define for any
distribution $T$ the product $f_l T$ via $\langle f_l T, \phi \rangle :=
\langle T, f_l \phi \rangle$ and it obviously follows $( f_l \delta) ( \phi) =
( f_l \phi) ( 0)$. If $f_l \phi$ is not in $C^{\infty}_0$ then it cannot be an
argument to $\delta \in \mathcal{D}'$ (although you can extend the domain of definition of $\delta$, of course).
In your example, and for some fixed test function, you can consider the delta
as the limit of bumps it is, and apply each of these to the product $f_l \phi$
then see what happens in the limit, which may not even exist depending on $f_l$ and $\phi$.
